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Showing posts from February, 2020

Crack The Code!

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Can you crack the 3 digit code?






Finding difficult? Here is the unlocking code!

Cracked The Code!

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What was the challenge?

Let's take a look at the given hints once again.




Let's name above numbers as NUMBER 1, NUMBER 2.....NUMBER 5 and hints as HINT 1, HINT 2......HINT 5.

NUMBER 1 and NUMBER 5 has one number common and that is 4 placed at same positions in both. The number 4 can't be the correct number as HINT 1 says it's correctly placed while Hint 5 suggests it is wrongly placed whereas in both numbers it's at same middle position.

So 4 can't be that CORRECT number.

Numbers 5,3 and 0 are eliminated by HINT 2. 

So numbers 5 and 4  are thrown out of race in NUMBER 1. Hence, the correct number pointed by HINT 1 must be 8 with it's correct position. Now, we got 1 digit of unlocking code XX8.

Now 5 also eliminated from NUMBER 3 leaving behind 1 and 7 as correct numbers pointed by HINT 3 but in wrong positions.

Since position of 1 is wrong and third digit of the code being already occupied correctly by 8 the 1 must be at second place of the code. And hence 7 must b…

"Go The Distance"

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There are 50 bikes with a tank that has the capacity to go 100 km. Using these 50 bikes, what is the maximum distance that you can go? 





Here is the maximum distance calculation!

Maximizing The Distance!

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What was the challenge?

Remember, there are 50 bikes, each with a tank that has the capacity to go 100 kms. 

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SOLUTION 1 :

Any body can think that these 50 bikes together can travel 50 x 100 = 5000 km. But this is not true in the case as all bikes will be starting from the same point. And we need to find how far we can we go from that point. 

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SOLUTION 2 :

Just launch all 50 bikes altogether from some starting point and go the distance of only 100 km with tanks of all bikes empty in the end.

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SOLUTION 3 :

1. Take all 50 bikes to 50 km so that tank of each is at half.

2. Pour fuels of 25 bikes (half filled) into other 25 bikes so that their tanks are full.

3. Now, move these 25 bikes to another 50 km so that again th…

So – who stole the apple?

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During lunch, 5 of Mr. Bryant’s students visit the supermarket.

One of the 5, stole an apple.

When questioned…

  Jim said: it was Hank or Tom.
Hank said: neither Eddie or I did it.
Tom Said: you’re both lying
Don said: no one of them is lying, the other is speaking the truth.
Eddie said: no Don, that’s not true.

When the shop owner asked Mr. Bryant, he said that three of the boys are always truthful, but two lie all the time.



So – who stole the apple?

And the name of the person who stole the apple is......!

Tom is an Apple Thief!

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What's the story behind the title?

Let's recall who said what.

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  Jim said: it was Hank or Tom.
Hank said: neither Eddie or I did it.
Tom Said: you’re both lying
Don said: no one of them is lying, the other is speaking the truth.
Eddie said: no Don, that’s not true.

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Clearly, Don and Eddie are making contradicting statements. Hence, one of them must be liar.

So there must be 1 more liar among Jim, Hank and Tom (since there are 2 liars & 3 truth tellers).

Tom's statement - you're both lying points Jim and Hank as liars. But there are total 2 liars with one being from either or Eddie as deduced above.

Hence, Tom himself must be that other liar.

Now we are sure that Jim and Hank must be telling the truth & as told by Hank, Eddie or Hank himself is not thief.

Since, Hank is not the one who stole the app…

Divide 1 Cube into 20 Cubes!

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From a 1987 Hungarian math contest for 11-year-olds:

How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?




Cut this way to get 20 cubes....

Division of 1 Cube into 20 cubes

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What was the challenge?

Mark cube for cutting 3 x 3 x 3 = 27 cubes. Cut any section of 2x2x2 = 8 cubes & cut rest of 27-8 = 19 cubes. So these 19 cubes plus 1 cube of 2x2x2 give us total number of 20 cubes. 



Need of Speed For Average Speed

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A man drives 1 mile to the top of a hill at 15 mph. How fast must he drive 1 mile down the other side to average 30 mph for the 2-mile trip?





Here is calculation of that speed needed!


Impossible Average Speed Challenge

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What was average speed challenge?

A man drives 1 mile to the top of a hill at 15 mph. That means he took, 1/15 hours i.e.4 minutes to reach at the top of a hill.

To achieve average speed of 30 mph, the man has to complete 2 miles trip in 1/15 hours i.e. 4 minutes. But he has already taken 4 minutes to reach at the top of a hill, hence he can't achieve average speed of 30 mph over entire trip. 



MATHEMATICAL PROOF:

Let 'x' be the speed needed in the journey down the hill.

Average Speed = Total Distance/Total time

Average Speed = (1 + 1)/(1/15 + 1/x)

30 = 2/(1/15 + 1/x)

A Visit To Grandmother's Home!

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A father wants to take his two sons to visit their grandmother, who lives 33 kilometers away. His motorcycle will cover 25 kilometers per hour if he rides alone, but the speed drops to 20 kph if he carries one passenger, and he cannot carry two. Each brother walks at 5 kph

Can the three of them reach grandmother’s house in 3 hours?



Do you think it's impossible? Click here!

Planning Journey Towards Grandmother's Home

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What was the challenge in the journey?

Yes, all three can reach at grandmother's home within 3 hours. Here is how.

Let M be the speed of motorcycle when father is alone, D be the speed of motorcycle when father is with son and S is speed of sons.  Let A and B are name of the sons.

As per data, M = 25 kph, D = 20 kph, S = 5 kph.

1. Father leaves with his first son A while asking second son B to walk. Father and A drives for 24 km in 24/20 = 6/5 hours. Meanwhile, son B walks (6/5) x 5 = 6 km.

2. Now father leaves down son A for walking and drives back to son B. The distance between them is 24 -6 = 18 km.





3. To get back to son B, father needs 18/(M+S) = 18/(25+5) = 18/30 = 3/5 hours & in that time son B walks for another (3/5) x 5 = 3 km. Now, son B is 6 + 3 = 9 km away from source where he meets his father. While son A walks another (3/5) x 5 = 3 km towards grandmother's home.


4. Right now father and B are 24 km while A is 6 km away from grandmother's home. So in another 24/20 …

Escape Safely to The Ground!

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You find yourself trapped at top an 800 foot tall building. The surrounding land is completely flat, plus there are no other structures nearby. You need to get to the bottom, uninjured, and can only safely fall about 5feet.

You look down the four walls; they are all completely smooth and featureless, except that one of the walls has a small ledge400feet above the ground. Furthermore, there are two hooks, one on this ledge, and one directly above it on the edge of the roof. The only tools you have are 600feet of rope, and a knife.

 How do you get to the bottom? 




This should be your strategy!

Strategy To Land Safely On The Ground

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Why strategy needed to be planned?

1.Tie one end of the rope to the to hook and climb down to the ledge.

2. Cut (without dropping) the rope that hangs below the ledge, then climb back to the roof carrying the extra rope that you cut. You now have two lengths of rope: one that is 400 feet long and one that is 200 feet long.

3.At the top, untie the rope from the hook.

 Now setup the ropes like : Tie a small loop at one end of the 200-foot long rope. String the 400-foot long rope through the loop so that half of its length is on either side of the loop. Make sure that the loop is snug enough that the 400-foot long rope won't fall out by itself, but loose enough that you can pull the rope out later.

4. Now, tie the end of the 200-foot rope without the loop to the first hook. The 200-foot long rope lets you climb halfway to the ledge. 

5.For the remaining 200 feet, you carefully climb down the 400-foot rope, which hangs down 200 feet from where it is held by the loop. 

6.Once you …

The Lightning Fast Addition!

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A  story tells that, as a 10-year-old schoolboy, Carl Friedrich Gauss was asked to find the sum of the first 100 integers. The tyrannical schoolmaster, who had intended this task to occupy the boy for some time, was astonished when Gauss presented the correct answer, 5050, almost immediately.


How did Gauss find it?

Actually, he used this trick!

Trick for The Lightneing Fast Addition!

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Why lightning fast speed needed?

Gauss attached 0 to the series and made pairs of numbers having addition 100.

100 + 0 = 100

99 + 1 = 100

98 + 2 = 100

97 + 3 = 100

96 + 4 = 100

95 + 5 = 100
..
..
..
..
..
..
51 + 49 = 100

This way he got 50 pair of integers (ranging in between 1-100) having sum equal to 100.

So sum of these 50 pairs = 100x50 = 5000.


And the number 50 left added to above total to get sum of integers 1 - 100 as 5000 + 50 = 5050

Story of 7 Generous Dwarfs

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The Seven Dwarfs are having breakfast, and Snow White has just poured them some milk. Before drinking, the dwarfs have a ritual. First, Dwarf #1 splits his milk equally among his brothers' mugs (leaving himself with nothing). Then Dwarf #2 does the same with his milk, etc. The process continues around the table, until Dwarf #7 has distributed his milk in this way. (Note that Dwarf #7 is named Dopey!) At the end, each dwarf has exactly the same amount of milk as he started with!
How much milk does each cup contain, if there were 42 ounces of milk altogether?

Finding difficult? Click here for answer!

Behind the Story of 7 Generous Dwarfs

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What was the story?

First thing is very clear that Dwarf 7 must have 0 ounces of milk at the start and end. Let's assume that 'a' be the maximum amount of milk (in ounces) that any dwarf has in his mug at any point of time. 

For a moment, let's assume Dwarf 1 himself has this 'a' amount of milk.



Now, when D1 distributes his 'a' amount of milk among 6 others, D7 receives 'a/6' amount of milk. At this point of time somebody else will be having maximum amount of milk 'a'. Let D2 be that person now having milk 'a'.


Next is D2's turn where he gives a/6 to all. So now D1 has a/6, D7 has 2a/6 and somebody else say D3 has maximum a. Continuing in this way, for each Dwarf's turn gives - 


Now, when we assumed D2 has maximum milk amount a after receiving a/6 from D1, then it's clear that he must had earlier 5a/6. Similarly, D3 had maximum amount of milk a after receiving a/6from D1 and D2 indicates that he had 4a/6 milk initially. Co…
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